5

我有一个包含 9 个图像的目录：

图像_0001、图像_0002、图像_0003
图像_0010，图像_0011
图像_0011-1、图像_0011-2、图像_0011-3
image_9999

我希望能够以有效的方式列出它们，如下所示（9 个图像的 4 个条目）：

(image_000[1-3], image_00[10-11], image_0011-[1-3], image_9999)

python中有没有办法以简短/清晰的方式返回图像目录（不列出每个文件）？

所以，可能是这样的：

列出所有图像，按数字排序，创建一个列表（从开始按顺序计算每个图像）。当图像丢失时（创建一个新列表），继续直到原始文件列表完成。现在我应该只有一些包含非损坏序列的列表。

我试图让阅读/描述数字列表变得容易。如果我有 1000 个连续文件的序列 它可以清楚地列为 file[0001-1000] 而不是 file['0001','0002','0003' 等...]

Edit1（基于建议）：给定一个扁平列表，您将如何推导出全局模式？

Edit2 我试图将问题分解成更小的部分。以下是部分解决方案的示例：data1 有效，data2 返回 0010 为 64，data3（实际数据）无效：

# Find runs of consecutive numbers using groupby.  The key to the solution
# is differencing with a range so that consecutive numbers all appear in
# same group.
from operator import itemgetter
from itertools import *

data1=[01,02,03,10,11,100,9999]
data2=[0001,0002,0003,0010,0011,0100,9999]
data3=['image_0001','image_0002','image_0003','image_0010','image_0011','image_0011-2','image_0011-3','image_0100','image_9999']

list1 = []
for k, g in groupby(enumerate(data1), lambda (i,x):i-x):
    list1.append(map(itemgetter(1), g))
print 'data1'
print list1

list2 = []
for k, g in groupby(enumerate(data2), lambda (i,x):i-x):
    list2.append(map(itemgetter(1), g))
print '\ndata2'
print list2

返回：

data1
[[1, 2, 3], [10, 11], [100], [9999]]

data2
[[1, 2, 3], [8, 9], [64], [9999]]
4

3 回答 3

6

这是您想要实现的工作的实现，使用您添加的代码作为起点：

#!/usr/bin/env python

import itertools
import re

# This algorithm only works if DATA is sorted.
DATA = ["image_0001", "image_0002", "image_0003",
        "image_0010", "image_0011",
        "image_0011-1", "image_0011-2", "image_0011-3",
        "image_0100", "image_9999"]

def extract_number(name):
    # Match the last number in the name and return it as a string,
    # including leading zeroes (that's important for formatting below).
    return re.findall(r"\d+$", name)[0]

def collapse_group(group):
    if len(group) == 1:
        return group[0][1]  # Unique names collapse to themselves.
    first = extract_number(group[0][1])  # Fetch range
    last = extract_number(group[-1][1])  # of this group.
    # Cheap way to compute the string length of the upper bound,
    # discarding leading zeroes.
    length = len(str(int(last)))
    # Now we have the length of the variable part of the names,
    # the rest is only formatting.
    return "%s[%s-%s]" % (group[0][1][:-length],
        first[-length:], last[-length:])

groups = [collapse_group(tuple(group)) \
    for key, group in itertools.groupby(enumerate(DATA),
        lambda(index, name): index - int(extract_number(name)))]

print groups

这打印出来['image_000[1-3]', 'image_00[10-11]', 'image_0011-[1-3]', 'image_0100', 'image_9999']，这就是你想要的。

历史：正如@Mark Ransom 在下面指出的那样，我最初回答了这个问题。为了历史，我最初的答案是：

您正在寻找glob。尝试：

import glob
images = glob.glob("image_[0-9]*")

或者，使用您的示例：

images = [glob.glob(pattern) for pattern in ("image_000[1-3]*",
    "image_00[10-11]*", "image_0011-[1-3]*", "image_9999*")]
images = [image for seq in images for image in seq]  # flatten the list
于 2010-10-13T20:39:16.040 回答
3

好的，所以我发现你的问题是一个有趣的谜题。我已经把如何“压缩”数字范围留给你了（标记为 TODO），因为有不同的方法来完成它，这取决于你喜欢它的格式以及你想要最少的元素数量还是最少的字符串描述长度。

此解决方案使用简单的正则表达式（数字字符串）将每个字符串分为两组：静态和变量。数据分类后，我使用groupby将静态数据收集成最长匹配组，达到汇总效果。我将整数索引前哨混合到结果中（在 matchGrouper 中），因此我可以从所有元素中重新选择不同的部分（在解包中）。

import re
import glob
from itertools import groupby
from operator import itemgetter

def classifyGroups(iterable, reObj=re.compile('\d+')):
    """Yields successive match lists, where each item in the list is either
    static text content, or a list of matching values.

     * `iterable` is a list of strings, such as glob('images/*')
     * `reObj` is a compiled regular expression that describes the
            variable section of the iterable you want to match and classify
    """
    def classify(text, pos=0):
        """Use a regular expression object to split the text into match and non-match sections"""
        r = []
        for m in reObj.finditer(text, pos):
            m0 = m.start()
            r.append((False, text[pos:m0]))
            pos = m.end()
            r.append((True, text[m0:pos]))
        r.append((False, text[pos:]))
        return r

    def matchGrouper(each):
        """Returns index of matches or origional text for non-matches"""
        return [(i if t else v) for i,(t,v) in enumerate(each)]

    def unpack(k,matches):
        """If the key is an integer, unpack the value array from matches"""
        if isinstance(k, int):
            k = [m[k][1] for m in matches]
        return k

    # classify each item into matches
    matchLists = (classify(t) for t in iterable)

    # group the matches by their static content
    for key, matches in groupby(matchLists, matchGrouper):
        matches = list(matches)
        # Yield a list of content matches.  Each entry is either text
        # from static content, or a list of matches
        yield [unpack(k, matches) for k in key]

最后，我们添加足够的逻辑来执行输出的漂亮打印，并运行一个示例。

def makeResultPretty(res):
    """Formats data somewhat like the question"""
    r = []
    for e in res:
        if isinstance(e, list):
            # TODO: collapse and simplify ranges as desired here
            if len(set(e))<=1:
                # it's a list of the same element
                e = e[0]
            else: 
                # prettify the list
                e = '['+' '.join(e)+']'
        r.append(e)
    return ''.join(r)

fnList = sorted(glob.glob('images/*'))
re_digits = re.compile(r'\d+')
for res in classifyGroups(fnList, re_digits):
    print makeResultPretty(res)

我的图像目录是根据您的示例创建的。您可以将 fnList 替换为以下列表进行测试：

fnList = [
 'images/image_0001.jpg',
 'images/image_0002.jpg',
 'images/image_0003.jpg',
 'images/image_0010.jpg',
 'images/image_0011-1.jpg',
 'images/image_0011-2.jpg',
 'images/image_0011-3.jpg',
 'images/image_0011.jpg',
 'images/image_9999.jpg']

当我针对这个目录运行时，我的输出看起来像：

StackOverflow/3926936% python classify.py
images/image_[0001 0002 0003 0010].jpg
images/image_0011-[1 2 3].jpg
images/image_[0011 9999].jpg
于 2010-10-13T23:03:56.910 回答
2 
def ranges(sorted_list):
    first = None
    for x in sorted_list:
        if first is None:
            first = last = x
        elif x == increment(last):
            last = x
        else:
            yield first, last
            first = last = x
    if first is not None:
        yield first, last

该increment功能留给读者作为练习。

编辑：这是一个如何使用整数而不是字符串作为输入的示例。

def increment(x): return x+1

list(ranges([1,2,3,4,6,7,8,10]))
[(1, 4), (6, 8), (10, 10)]

对于输入中的每个连续范围，您都会得到一对指示范围的开始和结束。如果元素不是范围的一部分，则开始值和结束值是相同的。

于 2010-10-13T21:18:34.323 回答